如何在嵌套函数中将所有可能的东西传递给 i、j 和 by？

Question

我正在开发一个内部使用 data.table 的包。在这个包中，我有一个函数 count_by，它按组计算 data.table 中特定变量的不同 ID 的数量。在一些帮助 () 的帮助下，我让它按预期工作：

library(data.table)
#> Warning: package 'data.table' was built under R version 3.6.2

# create example data
sample_dt <- data.table(
    id = sort(rep(LETTERS[1:3], 3)),
    year = rep(2018L:2020L),
    x = runif(9)
)
sample_dt[id == "B" & year < 2020, x := NA_real_]

# define inner function
count_by <- function(DT, id_var, val_var, by = NULL) {
    id_var <- as.character(substitute(id_var))
    val_var <- as.character(substitute(val_var))

    eval(substitute(
        DT[!is.na(get(val_var)), .(distinct_ids = uniqueN(get(id_var))), by = by]
    ))
}

# test inner function -> works as expected
(reference <- count_by(sample_dt, id_var = id, val_var = x, by = year))
#>    year distinct_ids
#> 1: 2018            2
#> 2: 2019            2
#> 3: 2020            3

identical(count_by(sample_dt, "id", x, year)       , reference)
#> [1] TRUE
identical(count_by(sample_dt, "id", "x", year)     , reference)
#> [1] TRUE
identical(count_by(sample_dt, "id", x, "year")     , reference)
#> [1] TRUE
identical(count_by(sample_dt, "id", x, c("year"))  , reference)
#> [1] TRUE
identical(count_by(sample_dt, "id", "x", "year")   , reference)
#> [1] TRUE
identical(count_by(sample_dt, "id", "x", c("year")), reference)
#> [1] TRUE
identical(count_by(sample_dt, id, "x", year)       , reference)
#> [1] TRUE
identical(count_by(sample_dt, id, "x", "year")     , reference)
#> [1] TRUE
identical(count_by(sample_dt, id, "x", c("year"))  , reference)
#> [1] TRUE
identical(count_by(sample_dt, id, x, "year")       , reference)
#> [1] TRUE
identical(count_by(sample_dt, id, x, c("year"))    , reference)
#> [1] TRUE

^{由 reprex package (v0.3.0)}

于 2020 年 2 月 20 日创建

现在我想在另一个函数中使用函数 count_by()（下面的最小示例）：

# define wrapper function
wrapper <- function(data, id_var, val_var, by = NULL) {
    data <- as.data.table(data)
    count_by(data, id_var, val_var, by)
}

# test wrapper function
wrapper(sample_dt, id_var = id, val_var = x, by = year)
#> Error in .(distinct_ids = uniqueN(get("id_var"))): could not find function "."

^{由 reprex package (v0.3.0)}

于 2020 年 2 月 20 日创建

调试 count_by() 导致观察到如果从 wrapper() 调用 count_by()，substitute(DT[...]) 也会将 DT 替换为 data：

Browse[2]> substitute(
+         DT[!is.na(get(val_var)), .(distinct_ids = uniqueN(get(id_var))), by = by]
+     )
data[!is.na(get("val_var")), .(distinct_ids = uniqueN(get("id_var"))), 
    by = by]

由于 data 在 count_by() 的函数环境中不可用，因此它被计算为 utils::data，然后导致错误。这使问题很清楚，但我想不出解决办法。

我需要替换整个 DT[...] 表达式以使 by 正常工作（参见或 将变量和名称传递给 data.table 函数)。但是我不能为了 DT 不被替换而替换整个表达式。

解决这个难题的方法是什么？

Answer 1

将 NSE 排除在外对于这个特定示例有效，并且大大简化了事情。但是你应该将参数作为字符串传递：

count_by <- function(DT, id_var, val_var, by = NULL) {
    DT[!is.na(get(val_var)), .(distinct_ids = uniqueN(get(id_var))), by = by]
}

wrapper <- function(data, id_var, val_var, by = NULL) {
    count_by(data, id_var, val_var, by)
}

wrapper(sample_dt, id_var = "id", val_var = "x", by = "year")

#    year distinct_ids
# 1: 2018            2
# 2: 2019            2
# 3: 2020            3

Answer 2

@chinsoon12，非常感谢！你的回答几乎做到了！我仍然需要将 id_var 和 val_var 转换为字符，然后在对 data.table 的调用中将 get() 这些转换为字符 - 否则，将字符串传递给 id_var 和 val_var 不起作用。

但是在更高层次上进行评估是必要的关键思想。以下是供日后参考的完整答案：

# define inner function
count_by <- function(DT, id_var, val_var, by = NULL) {
    id_var <- as.character(substitute(id_var))
    val_var <- as.character(substitute(val_var))

    substitute(
        DT[!is.na(get(val_var)), .(distinct_ids = uniqueN(get(id_var))), by = by]
    )
}

# define wrapper function
wrapper <- function(data, id_var, val_var, by = NULL) {
    data <- as.data.table(data)
    expr <- eval(substitute(count_by(data, id_var, val_var, by)))
    eval(expr)
}

# test wrapper function
(reference <- (wrapper(sample_dt, id_var = id, val_var = x, by = year)))
#>    year distinct_ids
#> 1: 2018            2
#> 2: 2019            2
#> 3: 2020            3

identical(wrapper(sample_dt, "id", x, year)       , reference)
#> [1] TRUE
identical(wrapper(sample_dt, "id", "x", year)     , reference)
#> [1] TRUE
identical(wrapper(sample_dt, "id", x, "year")     , reference)
#> [1] TRUE
identical(wrapper(sample_dt, "id", x, c("year"))  , reference)
#> [1] TRUE
identical(wrapper(sample_dt, "id", "x", "year")   , reference)
#> [1] TRUE
identical(wrapper(sample_dt, "id", "x", c("year")), reference)
#> [1] TRUE
identical(wrapper(sample_dt, id, "x", year)       , reference)
#> [1] TRUE
identical(wrapper(sample_dt, id, "x", "year")     , reference)
#> [1] TRUE
identical(wrapper(sample_dt, id, "x", c("year"))  , reference)
#> [1] TRUE
identical(wrapper(sample_dt, id, x, "year")       , reference)
#> [1] TRUE
identical(wrapper(sample_dt, id, x, c("year"))    , reference)
#> [1] TRUE

# test expression in by
wrapper(sample_dt, "id", x, by = .(year_2019 = year > 2019L))
#>    year_2019 distinct_ids
#> 1:     FALSE            2
#> 2:      TRUE            3

^{由 reprex package (v0.3.0)}

于 2020 年 2 月 20 日创建

如何在嵌套函数中将所有可能的东西传递给 i、j 和 by？

How to pass everything possible to i, j and by in nested functions?

r

data.table

non-standard-evaluation